Question

If $a{x}^4+b{x}^3+c{x}^2+dx+e$ = $\left|\begin{array}{ccc}{x}^3+3x& x-1& x+3\\ x+1& -2x& x-4\\ x-3& x+4& 3x\end{array}\right|$ then show $e=0$.

Answer 1

Now e is obtained by putting x=0 in L.H.S.

Hence putting x=0 on both sides of the given equation,

We get e= $\left|\begin{array}{ccc}0& -1& 3\\ 1& 0& -4\\ -3& 4& 0\end{array}\right|$= 0 being a skew-symmetric determinant of odd order.